# Thread: Prove the sum of a harmonic series is never an integer

1. ## Prove the sum of a harmonic series is never an integer

Not really sure how to do this. I thought about breaking it up into the numerator and denominator with a denominator of n!, but I don't know how to create a closed form expression of the numerator.

Either way, I don't think that's quite the right approach.

2. Originally Posted by davismj

Not really sure how to do this. I thought about breaking it up into the numerator and denominator with a denominator of n!, but I don't know how to create a closed form expression of the numerator.

Either way, I don't think that's quite the right approach.

Let $\displaystyle k\in \mathbb{N}$ be s.t. $\displaystyle 2^k\le n <2^{k+1}$ and let us define $\displaystyle L:=lcm\{2,3,...,n\}=2^k3^a5^b\cdot\ldots$ , so:

$\displaystyle \sum\limits_{i=2}^n\frac{1}{i} = \frac{(2^k3^a\cdot\ldots)\slash 2}{L}+\frac{(2^k3^a\cdot\ldots)\slash 3}{L}+\ldots +\frac{(2^k3^a\cdot\ldots)\slash n}{L}$

Now pay attention to the fact that all the numerators above are even except precisely one (why?), whereas the common denominator is even, and this mean this number can't be an integer...

Tonio

3. Originally Posted by tonio
Let $\displaystyle k\in \mathbb{N}$ be s.t. $\displaystyle 2^k\le n <2^{k+1}$ and let us define $\displaystyle L:=lcm\{2,3,...,n\}=2^k3^a5^b\cdot\ldots$ , so:

$\displaystyle \sum\limits_{i=2}^n\frac{1}{i} = \frac{(2^k3^a\cdot\ldots)\slash 2}{L}+\frac{(2^k3^a\cdot\ldots)\slash 3}{L}+\ldots +\frac{(2^k3^a\cdot\ldots)\slash n}{L}$

Now pay attention to the fact that all the numerators above are even except precisely one (why?)
Because when you get to the $\displaystyle (2^k-1)th$ term, $\displaystyle (2^k3^a\ldots)/2^k$ reduces to the product of the remaining primes, all odd.

Thank you muchly. I see it now.

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# prove that 1 1 2 1 3 1 n is not an integer

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